Cho hàm số \(y = f\left( x \right) = \sqrt {\tan x + \cot x} \). Giá trị \(f'\left( {\frac{\pi }{4}} \right)\) bằng:

Hướng dẫn giải:

Chọn C.

\(y = \sqrt {\tan x + \cot x} \Rightarrow {y^2} = \tan x + \cot x \Rightarrow y'.2y = \frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}\).

\( \Rightarrow y' = \frac{1}{{2\sqrt {\tan x + \cot x} }}\left( {\frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right)\).

\(f'\left( {\frac{\pi }{4}} \right) = \frac{1}{{2\sqrt {\tan \frac{\pi }{4} + \cot \frac{\pi }{4}} }}\left( {\frac{1}{{{{\cos }^2}\left( {\frac{\pi }{4}} \right)}} - \frac{1}{{{{\sin }^2}\left( {\frac{\pi }{4}} \right)}}} \right) = \frac{1}{{2\sqrt 2 }}\left( {2 - 2} \right) = 0\)